Redox Reactions

${{\mathrm{ClO}}_4}^{-}$ does not undergo disproportionation reaction because chlorine is present in it's highest oxidation state i.e. +7.

(2)

$3{\mathrm{Cl}}_2+6{\mathrm{OH}}^{-}\to 5{\mathrm{Cl}}^{-}+{\mathrm{ClO}}_3^{-}+3{\mathrm{H}}_2\mathrm{O}$

BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl is a double displacement reaction with no change in oxidation states, so it is not a redox reaction. In the other options, there is a clear change in oxidation numbers indicating redox processes.

${{\mathrm{ClO}}_4}^{-}$ → chlorine is in its highest oxidation state i.e., +7.
Chlorine can exhibit a –1 to +7 oxidation state.
The oxidation states of chlorine which can undergo disproportionation are : 0, +1, +3, and +5.

Oxidising agent wahi ban sakta hai jo khud reduce ho sake. N³⁻ me nitrogen ki oxidation state −3 hoti hai jo minimum hoti hai, aur ye aur reduce nahi ho sakta. Baaki ions (SO₄²⁻, BrO₃⁻, MnO₄⁻) me central atom ki high oxidation state hoti hai jo reduce ho sakti hai, isliye wo oxidising agent ho sakte hain.

Correct answer: N³⁻ cannot function as oxidising agent.

Mohr’s salt (ferrous ammonium sulphate) is commonly dissolved in slightly acidic solution (using dilute H₂SO₄) to prevent oxidation and hydrolysis of Fe²⁺ ions. The acidic medium suppresses Fe²⁺→Fe³⁺ conversion and formation of insoluble iron hydroxides, ensuring the stable, clear solution of Mohr’s salt.

${{\mathrm{ClO}}_4}^{-}$ → chlorine is in its highest oxidation state i.e., +7.
Chlorine can exhibit a –1 to +7 oxidation state.
The oxidation states of chlorine which can undergo disproportionation are : 0, +1, +3, and +5.

The reaction is

${\mathrm{MnO}}_4^{-}$ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.

Using N₁V₁ = N₂V₂,

$\frac{1}{8}\times 5V_1=\frac{1}{4}\times 25$

Solving, V₁ = 10 mL.

Correct explanation: Bromine is a stronger oxidising agent than iodine, isliye Br₂ thiosulphate ko oxidise karta hai jabki I₂ usse reduce hota hai. Isliye thiosulphate ka behaviour Br₂ aur I₂ ke saath alag hota hai.

Correct option: C – Bromine is a stronger oxidant than iodine.

In the oxalic acid $\left(H_2{C}_2{O}_4\right)$ vs $KMn{O}_4$ titration, the oxidation of oxalate $\left(C_2{O_4}^{2-}\right)$ to $C{O}_2$ is slow at room temperature.

$5C_2{O}_4^{2-}+2Mn{O}_4^{-}+16H^{+}\to 10C{O}_2+$
$2M{n}^{2+}+8H_{2}O$
Why heating is needed?
- The initial formation of $M{n}^{2+}$ from $KMn{O}_4$ is slow at room temperature.
- $M{n}^{2+}$ formed acts as a catalyst, accelerating the rest of the reaction.
- Hence, the solution is heated to around $60°C$ to speed up the reaction.
In $FAS\left(F{e}^{2+}\right)$ vs $KMn{O}_4$ titration, the reaction proceeds quickly at room temperature:

$5F{e}^{2+}+Mn{O}_4^{-}+8H^{+}\to 5F{e}^{3+}+M{n}^{2+}+4H_{2}O$

Why no heating is needed?
- The oxidation of $F{e}^{2+}$ is already fast at room temperature.
- Heating can cause $F{e}^{2+}$ to be oxidized to $F{e}^{3+}$ by air, leading to errors.
Hence, both statements are true.

Balanced redox equation in alkaline medium:

Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O

To balance:

Unbalanced form given in question:

aCl₂ + bOH⁻ → cClO⁻ + dCl⁻ + eH₂O

Balanced form:

a = 1, b = 2, c = 1, d = 1, e = 1

Hence, the correct option is: A (1, 2, 1, 1).

• (a) $\mathrm{Fe^{3+}}$ oxidises $\mathrm{I^-}$ to $\mathrm{I_2}$ and itself gets reduced to $\mathrm{Fe^{2+}}$ → correct redox reaction.

• (b) $\mathrm{FeI_3}$ is not stable in aqueous solution → incorrect.

• (c) $\mathrm{Fe^{2+}}$ cannot be reduced to Fe (0) by $\mathrm{S_2O_8^{2-}}$ → incorrect.

• (d) $\mathrm{S_2O_8^{2-}}$​ is a strong oxidant and oxidises $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$ → correct.